Assume we work (over $\mathbb{C}$) on a polarized K3 surface $(X,L)$ with a line bundle $M$ on $X$ such that $M^2=-6$ and $ML=0$ as well as $h^0(M)=h^2(M)=0$ and thus $h^1(M)=1$.
Then $E=\mathcal{O}_X\oplus M$ is $\mu_L$-semistable and $v(E)=(2,M,-1)$ so that $v(E)^2=-2$, hence $v=v(E)$ is an exceptional Mukai vector. We find $ext^i(E,E)=2$ for $i=0,1,2$.
According to a result of Kuleshov the Muaki vector $v$ can also be realized by a $\underline{\mathrm{simple}}$ $\mu_L$-semistable bundle $F$, that is we have $v(F)=v$ and $ext^i(F,F)=1$ for $i=0,2$ and $ext^1(F,F)=0$. I think $F$ is given by a nontrivial extension $0 \rightarrow \mathcal{O}_X \rightarrow F \rightarrow M\rightarrow 0$.
$\textbf{Question:}$ How does the moduli space $M_L(v)$ of semistable sheaves with Mukai vector $v$ look in this case?
It cannot contain stable sheaves, because then by a result of Mukai it has to be a reduced point and all bundles are isomorphic, but we have constructed an explicit strictly semistable bundle $E$ in it. So is it a non-reduced point and the point is given by the S-equivalence class of $E$? So that $F$ and $E$ define the same point in $M_L(v)$?
How to compute the tangent space $T_EM_L(v)$ in this case? I think on curves the tangent space at a semistable bundle of the form $G_1\oplus G_2$ is given by $Ext^1(G_1,G_1)\oplus Ext^1(G_1,G_2)\oplus Ext^1(G_1,G_2)\otimes Ext^1(G_2,G1)$. Is such a description also valid on surfaces? I think on curves one uses Luna's slice theorem and the smoothness of the Quot scheme to compute the tangent space. I am not sure this can be done on a surface.
